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I want to show that $\langle x,y \mid x^3=y^2=(xy)^3=e\rangle$ is isomorphic to $\mathfrak{A}_4$. I've already done most of the proof. The remaining and probably hardest part is to show that $\langle x,y \mid x^3=y^2=(xy)^3=e\rangle$ has at most $12$ elements.

For this I have a hint: Show that the conjugates of $y$, together with the identity element form a normal subgroup of order $\leq 4$ and index $\leq 3$.

I suspect that the conjugates of $y$ are just $y$, $xyx^{-1}$ and $x^2yx^{-2}$, but I really don't know how to prove this. Can somebody help me?

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    $\begingroup$ Hint: Write $y^x$ for $xyx^{-1}$, etc. Then $H = \langle y,y^x \rangle$ is a dihedral group, since it is generated by a pair of elements of order $2.$ Then we have $y.y^{x}.y^{x^{2}} = (yx)^3 =1$ since $yx$ is conjugate to $xy.$ Thus $H = \langle y,y^x,y^{x^{2}} \rangle.$ Then $H$ is normalized both by $y$ and by $x$ $\endgroup$ – Geoff Robinson Feb 24 '12 at 8:43
  • $\begingroup$ @Geoff: Thanks for the hint. I think I can do the proof from here. I'll write an answer then. $\endgroup$ – Stefan Feb 25 '12 at 11:33
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It is easy to show that $(123)$ and $(12)(34)$ generate $\mathfrak{A}_4$ and satisfy the given relations with $(123)$ in place of $x$ and $(12)(34)$ in place of $y$. So there exists a group homomorphism of $G:=\langle x,y|x^3=y^2=(xy)^3=e\rangle$ onto $\mathfrak{A}_4$. If we can show that $G$ cannot have more than $12$ elements, the proof is finished.

We have $$y(xyx^{-1})(x^2yx^{-2})=(yx)^3x^{-3}=(yx)^3=(y(xy)y^{-1})^3=(xy)^3=e.$$ Since $y$ is equal to its own inverse, the same is true for its conjugates. This information allows us to play around with the equation above to conclude that $H:=\{e,y,xyx^{-1},x^2yx^{-2}\}$ is indeed a subgroup.

There exists a homomorphism $f$ of $G$ onto $\mathbf{Z}/3\mathbf{Z}$ mapping $x$ to $1$ and $y$ to $0$. We show that $\text{Ker}(f)=H$ and thereby $[G:H]=3$, which implies $|G|\leq 12$.

Let $\phi$ be the homomorphism of the free monoid $\text{Mo}(\xi,\eta)$ onto $G$ mapping $\xi$ to $x$ and $\eta$ to $y$. Then $\phi(w)\in\text{Ker}(f)$ if and only if the number of occurences of $\xi$ in $w$ is a multiple of $3$. For $i\in\mathbf{N}$, let $M_i$ be the set of elements of $\text{Mo}(\xi,\eta)$ with $i$ occurences of $\xi$. We show by induction on $n$ that $\phi(M_{3n})\subset H$ for all $n\in\mathbf{N}$.

Obviously, $\phi(M_0)=\{e,y\}\subset H$. If $n\geq 1$ and $w\in M_{3n}$, then there exists $w'\in M_3$ and $w''\in M_{3(n-1)}$ such that $w=w'w''$. Since $\phi(w'')\in H$ by induction, it is left to show that $\phi(M_3)\subset H$. Let $w\in M_3$. Then $w=\eta^i\xi\eta^j\xi\eta^k\xi\eta^l$ with $i,j,k,l\in\mathbf{N}$. So $\phi(w)=y^ixy^jxy^kxy^l$. It remains to show that $xy^jxy^kx\in H$ for all $j,k\in\mathbf{N}$.

There are four cases:

$j$ and $k$ are even $\Rightarrow$ $xy^jxy^kx=x^3=e\in H$.

$j$ is even and $k$ is odd $\Rightarrow$ $xy^jxy^kx=x^2yx=x^2yx^{-2}\in H$

$j$ is odd and $k$ is even $\Rightarrow$ $xy^jxy^kx=xyx^2=xyx^{-1}\in H$

$j$ and $k$ are odd $\Rightarrow$ $xy^jxy^kx=xyxyx=(xy)^3y=y\in H$.

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